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Evaluate the following : ∫t.sin-1t1-t2.dt - Mathematics and Statistics

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प्रश्न

Evaluate the following : `int (t.sin^-1 t)/sqrt(1 - t^2).dt`

योग
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उत्तर

Let I = `int (t.sin^-1 t)/sqrt(1 - t^2).dt`

= `int t.sin^-1 t. 1/sqrt(1 - t^2).dt`

Put sin–1 t = θ

∴ `1/sqrt(1 - t^2).dt` = dθ
and
t = sin θ
∴ I = `int (sinθ).θdθ`

= `int θ sin θ dθ`

= `θ int sin θ dθ - int [d/(dθ) (θ) int sin θ dθ]dθ`

= `θ (- cos θ) - int 1. (- cosθ)dθ`

= `- θ cosθ + int cosθ  dθ`

= – θ cos θ + sin θ + c

= `- θ.sqrt(1 - sin^2θ) + sin θ + c`

= `- sin^-1 t.sqrt(1 - t^2) + t + c`

= `- sqrt(1 - t^2).sin^-1 t + t + c`.

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Indefinite Integration - Exercise 3.3 [पृष्ठ १३७]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 3 Indefinite Integration
Exercise 3.3 | Q 1.14 | पृष्ठ १३७

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